If $Y\cap \rm int(B)\neq\emptyset$ then $Y\subset [\rm int(B\setminus \rm int(Y))]^c$?

Question

I have two closed sets $Y$ and $B$ from a Banach space $X$

I need to prove that if $Y\cap \rm int(B)\neq \emptyset $ then $$Y\subset [\rm int(B\setminus \rm int(Y))]^c$$

I started like this:

Let $u\in Y$

$\bullet$ If $u\in \rm int(Y)$ then $u\notin B\setminus \rm int(Y)$

$\bullet$ If $ u\not\in \rm int(Y)$(that is $u\in \partial (Y)$ )

If i suppose by contradiction that $u\in \rm int(B\setminus \rm int(Y))$ then $B\setminus \rm int(Y)$ is a neighborhood of $u$

as $u\in Y=\overline{Y}$ $$ (B\setminus \rm int(Y))\cap Y\neq\emptyset $$

but i don't find contradiction

How to do please?

Thank you

Answer 1

Let Y = B = [0,1]. B - int Y = {0,1}.
$Y \subset (B - int Y)^c$ is not possible.
Notice that the title is inconsistent with your question.

Let X = R, Y = {1}, B = [0,2].
Is 1 in R - (int B - int Y)?